Q103 / 160 · AP EAMCET 2023 (18 May Shift 2) · Mathematics
$f(x)$ and $g(x)$ are differentiable functions such that $\frac{f(x)}{g(x)}=$ a non zero constant. If $\frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{g}^{\prime}(\mathrm{x})}=\alpha(\mathrm{x})$ and $\left(\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}\right)^{\prime}=\beta(\mathrm{x})$, then $\frac{\alpha(x)-\beta(x)}{\alpha(x)+\beta(x)}=$
- A0
- B$f(x)+g(x)$
- Correct answerC1
- D$\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{g}^{\prime}(\mathrm{x})$
Solution
$$
\begin{aligned}
& \text {} \because \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{g}^{\prime}(\mathrm{x})}=\alpha(\mathrm{x}) \&\left(\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}\right)^{\prime}=\beta(\mathrm{x}) \\
& \because \frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}=\text { a non-zero constant, } \\
& \Rightarrow\left(\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}\right)^{\prime}=0 \Rightarrow \beta(\mathrm{x})=0
\end{aligned}
$$
Now, $\frac{\alpha(x)-\beta(x)}{\alpha(x)+\beta(x)}=\frac{\alpha(x)-0}{\alpha(x)+0}=1$
Hence, the correct answer is option: 'C'.